May 2015 Mathematics Higher level Paper 2 - IB Documents

M15/5/MATHL/HP2/ENG/TZ2/XX/M

22 pages

Markscheme

May 2015

Mathematics

Higher level

Paper 2

– 2 – M15/5/MATHL/HP2/ENG/TZ2/XX/M

This markscheme is confidential and for the exclusive use of examiners in this examination session. It is the property of the International Baccalaureate and must not be reproduced or distributed to any other person without the authorization of the IB Assessment Centre.


Instructions to Examiners

Abbreviations M Marks awarded for attempting to use a valid Method; working must be seen. (M) Marks awarded for Method; may be implied by correct subsequent working.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

(A) Marks awarded for an Answer or for Accuracy; may be implied by correct subsequent working. R Marks awarded for clear Reasoning. N Marks awarded for correct answers if no working shown. AG Answer given in the question and so no marks are awarded.

Using the markscheme 1 General

Mark according to RM™ Assessor instructions and the document “Mathematics HL: Guidance for e-marking May 2015”. It is essential that you read this document before you start marking. In particular, please note the following. Marks must be recorded using the annotation stamps. Please check that you are entering marks for the right question. If a part is completely correct, (and gains all the ‘must be seen’ marks), use the ticks with

numbers to stamp full marks. If a part is completely wrong, stamp A0 by the final answer. If a part gains anything else, it must be recorded using all the annotations. All the marks will be added and recorded by RM™ Assessor.

2 Method and Answer/Accuracy marks

Do not automatically award full marks for a correct answer; all working must be checked, and marks awarded according to the markscheme.

It is not possible to award M0 followed by A1, as A mark(s) depend on the preceding M mark(s), if any.

Where M and A marks are noted on the same line, for example, M1A1, this usually means M1 for an attempt to use an appropriate method (for example, substitution into a formula) and A1 for using the correct values.

Where the markscheme specifies (M2), N3, etc, do not split the marks.


Once a correct answer to a question or part-question is seen, ignore further correct working. However, if further working indicates a lack of mathematical understanding do not award the final A1. An exception to this may be in numerical answers, where a correct exact value is followed by an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part, and correct FT working shown, award FT marks as appropriate but do not award the final A1 in that part.

Examples

Correct answer seen Further working seen Action 1.

8 2 5.65685... (incorrect decimal value)

Award the final A1 (ignore the further working)

2. 1 sin 44

x sin x Do not award the final A1

3. log loga b log ( )a b Do not award the final A1

3 N marks Award N marks for correct answers where there is no working.

Do not award a mixture of N and other marks. There may be fewer N marks available than the total of M, A and R marks; this is deliberate as it

penalizes candidates for not following the instruction to show their working.

4 Implied marks Implied marks appear in brackets, for example, (M1), and can only be awarded if correct work is

seen or if implied in subsequent working. Normally the correct work is seen or implied in the next line. Marks without brackets can only be awarded for work that is seen.

5 Follow through marks Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s). To award FT marks, there must be working present and not just a final answer based on an incorrect answer to a previous part. If the question becomes much simpler because of an error then use discretion to award fewer

FT marks. If the error leads to an inappropriate value (for example, sin 1.5 ), do not award the mark(s)

for the final answer(s). Within a question part, once an error is made, no further dependent A marks can be awarded,

but M marks may be awarded if appropriate. Exceptions to this rule will be explicitly noted on the markscheme.


6 Mis-read

If a candidate incorrectly copies information from the question, this is a mis-read (MR). A candidate should be penalized only once for a particular mis-read. Use the MR stamp to indicate that this has been a misread. Then deduct the first of the marks to be awarded, even if this is an M mark, but award all others so that the candidate only loses one mark. If the question becomes much simpler because of the MR, then use discretion to award fewer

marks. If the MR leads to an inappropriate value (for example, sin 1.5 ), do not award the mark(s) for

the final answer(s). 7 Discretionary marks (d) An examiner uses discretion to award a mark on the rare occasions when the markscheme does

not cover the work seen. In such cases the annotation DM should be used and a brief note written next to the mark explaining this decision.

8 Alternative methods

Candidates will sometimes use methods other than those in the markscheme. Unless the question

specifies a method, other correct methods should be marked in line with the markscheme. If in doubt, contact your team leader for advice.

Alternative methods for complete questions are indicated by METHOD 1, METHOD 2, etc. Alternative solutions for part-questions are indicated by EITHER . . . OR. Where possible, alignment will also be used to assist examiners in identifying where these

alternatives start and finish. 9 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

As this is an international examination, accept all alternative forms of notation. In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer. In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the bracket or the form in brackets (if it is seen).

Example: for differentiating ( ) 2sin (5 3)f x x , the markscheme gives: ( ) 2cos (5 3) 5f x x 10cos(5 3)x A1

Award A1 for 2cos (5 3) 5x , even if 10cos (5 3)x is not seen.


10 Accuracy of Answers Candidates should NO LONGER be penalized for an accuracy error (AP). If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to

the required accuracy. When this is not specified in the question, all numerical answers should be given exactly or correct to three significant figures. Please check work carefully for FT.

11 Crossed out work

If a candidate has drawn a line through work on their examination script, or in some other way crossed out their work, do not award any marks for that work.

12 Calculators

A GDC is required for paper 2, but calculators with symbolic manipulation features (for example, TI-89) are not allowed.

Calculator notation The Mathematics HL guide says: Students must always use correct mathematical notation, not calculator notation.

Do not accept final answers written using calculator notation. However, do not penalize the use of calculator notation in the working.

13 More than one solution

Where a candidate offers two or more different answers to the same question, an examiner should only mark the first response unless the candidate indicates otherwise.

14. Candidate work

Candidates are meant to write their answers to Section A on the question paper (QP), and Section B on answer booklets. Sometimes, they need more room for Section A, and use the booklet (and often comment to this effect on the QP), or write outside the box. This work should be marked. The instructions tell candidates not to write on Section B of the QP. Thus they may well have done some rough work here which they assume will be ignored. If they have solutions on the answer booklets, there is no need to look at the QP. However, if there are whole questions or whole part solutions missing on answer booklets, please check to make sure that they are not on the QP, and if they are, mark those whole questions or whole part solutions that have not been written on answer booklets.


Section A

1. (a) 1 5 12 sin1002

A (M1)

229.5 cm A1

[2 marks]

(b) 2 2 2AC 5 12 2 5 12 cos100 (M1)

therefore AC 13.8 (cm) A1 [2 marks]

Total [4 marks]

2. (a)

11 11 10 9 8 3304 4 3 2 1

(M1)A1

[2 marks]

(b) 5 6 5 4 6 52 2 2 1 2 1

M1

150 A1

[2 marks]

(c) METHOD1

number of ways all men5

54

330 5 325 M1A1

Note: Allow FT from answer obtained in part (a).

[2 marks]

continued…


Question 2 continued.

METHOD 2

6 5 6 5 6 5 61 3 2 2 3 1 4

M1

=325 A1

[2 marks]

Total [6 marks]

3. (a)

general shape including 2 minimums, cusp A1A1 correct domain and symmetrical about the middle ( 5)x A1

[3 marks]

(b) 9.16x or 0.838x A1A1

[2 marks]

Total [5 marks]


4. (a) (i) ~ (5)X PoP( 8) 0.133X (M1)A1

(ii) 7 0.133 M1 0.934 days A1

Note: Accept “1 day”.

[4 marks]

(b) 7 5 35 35Y ~ Po( ) (A1)

P( 29) 0.177Y (M1)A1

[3 marks]

Total [7 marks]

5. (a)

2(0) 2 22 1(0) 2

2 2

b ba ab a b a

u v (M1)(A1)

2 422

ba b

b a c

(M1)

1, 2, 4a b c A2

Note: Award A1 for two correct.

[5 marks]

(b)

424

n (A1)

4 2 4 0x y z (2 2 0)x y z A1 [2 marks]

Total [7 marks]


6. (a) EITHER

ln ( ) ln (5 10)y x a b x (M1)

ln ( ) ln ln (5 10)y x a c x

ln ( ) ln (5 10)y c x a x (M1)

OR

ln (5 10) ln 5( 2)y x x (M1)

ln (5) ln ( 2)y x (M1)

THEN

2, ln 5a b A1A1

Note: Accept graphical approaches.

Note: Accept a 2, b 1.61

[4marks]

(b) 2 2π ln (5 10) de

eV x x (M1)

99.2 A1

[2marks]

Total [6 marks]


7. (a) 2 6 0x y z

4 3 14 4x y z

2 2 ( 2) 12x y z

attempt at row reduction M1

eg 2 12R R and 3 1R R

2 6 0x y z

2 4y z

3 ( 8) 12y z A1

eg 3 23R R

2 6 02 4

( 2)

x y zy z

z

A1

(i) no solutions if 2 , 0 A1

(ii) one solution if 2 A1

(iii) infinite solutions if 2 , 0 A1

Note: Accept alternative methods e.g. determinant of a matrix

Note: Award A1A1A0 if all three consistent with their reduced form, A1A0A0 if two or one answer consistent with their reduced form.

[6 marks]

(b) 2 4 4 2y z y z A1

2 6 2 4 6 4 4 2 2x y z z z z x z A1

therefore Cartesian equation is 2 4

2 2 1x y z

or equivalent A1

[3 marks]

Total [9 marks]


8. (a)

EITHER

area of triangle 1 3 4 ( 6)2

A1

area of sector 21 4arcsin 5 ( 11.5911 )2 5

A1

OR

42

0

25 dx x M1A1

THEN

total area 17.5911 m2 (A1)

percentage 17.5911 100 44%

40

A1

[4 marks] (b) METHOD 1

area of triangle 21 4 162

a A1

4arcsina

(A1)

area of sector 2 21 1 4arcsin2 2r a

a

A1

therefore total area 2 21 42 16 arcsin 202

a aa

A1

rearrange to give: 2 24arcsin 4 16 40a aa

AG

continued…


Question 8 continued

METHOD 2

42 2

0

20a x dx M1

use substitution dsin , cosdxx a a

4arcsin

2 2

0

cos d 20a

a

4arcsin2

0

cos 2 1 d 202

aa

M1

4arcsin2

0

sin 2 402

aa

A1

4arcsin2

0sin cos 40aa

22 24 4 4arcsin 1 40a a

a a a

A1

2 24arcsin 4 16 40a aa

AG

[4 marks]

(c) solving using GDC 5.53cma A2 [2 marks]

Total [10 marks]


9.

(a) attempt to set up the problem using a tree diagram and/or an equation,

with the unknown x M1

4 2 13(1 )5 3 18x x A1

4 2 13 25 3 18 3x x

2 115 18x

5

12x A1

[3 marks] (b) attempt to set up the problem using conditional probability M1 EITHER

5 112 5

13118

A1

OR

5 112 51 7

12 36

A1

THEN

3

10 A1

[3 marks] Total [6 marks]


Section B

10. (a) (i) (110 130)P X 0.49969... 0.500 50.0% (M1)A1

Note: Accept 50

Note: Award M1A0 for 0.50 (0.500)

(ii) ( 130) (1 0.707 ) 0.293P X M1

expected number of turnips 29.3 A1

Note: Accept 29.

(iii) no of turnips weighing more than 130 is ~ (100, 0.293)Y B M1

( 30) 0.478P Y A1

[6 marks]

(b) (i) 2~ 144,X N 1( 130) 0.0667

15P X (M1)

130 144 0.0667P Z

14 1.501

(A1)

9.33g A1

(ii)( 150)( 150 | 130)( 130)P XP X XP X

M1

0.26008 0.2791 0.06667

A1

expected number of turnips 55.7 A1

[6 marks]

Total [12 marks]


11. (a) attempt at implicit differentiation M1

d d2 5 5 2 0d dy yx x y yx x

A1A1

Note: A1 for differentiation of 2 5x xy , A1 for differentiation of 2y and 7.

d2 5 (2 5 ) 0dyx y y xx

d 5 2d 2 5y y xx y x

AG

[3 marks]

(b) d 5 1 2 6 1d 2 1 5 6 4yx

A1

gradient of normal 4 A1

equation of normal 4y x c M1

substitution of (6, 1)

4 25y x A1

Note: Accept 1 4( 6)y x

[4 marks]

(c) setting 5 2 12 5y xy x

M1

y x A1 substituting into original equation M1

2 2 25 7x x x (A1)27 7x

1x A1 points (1, 1) and ( 1, 1) (A1)

distance 8 2 2 (M1)A1

[8 marks]

Total [15 marks]


12. (a) METHOD 1

2 2 39s 9 3 d ( )2

t t t t t c (M1)

0, 3 3t s c (A1)

4 11t s A1 [3 marks]

METHOD 2

4

2

0

3 9 3s t t dt (M1)(A1)

11s A1 [3 marks]

(b) 2 3932

s t t (A1)

correct shape over correct domain A1 maximum at (3, 16.5) A1

t intercept at 4.64, s intercept at 3 A1 minimum at (5, 9.5) A1

[5 marks] continued…


Question 12 continued

(c) 9.5 cos 2a b

16.5 cos3a b (M1)

Note: Only award M1 if two simultaneous equations are formed over the correct domain.

72

a A1

13b A1 [3 marks]

(d) at 1t :

2 393 32t t (M1)

2 9 02

t t

192

t A1

solving 7 213cos 32 5

t (M1)

2GDC 6.22t A1

Note: Accept graphical approaches.

[4 marks] Total [15 marks]


13. (a) 1L and 2L are not parallel, since

1 21 12 6

k

R1

if they meet, then 1 1 2 and 2 2 M1

solving simultaneously 0 A1

2 2 4 6 2 4 contradiction, R1 so lines are skew AG

Note: Do not award the second R1 if their values of parameters are incorrect.

[4 marks]

(b) 1 2

1 1 11 6 41cos2 6

M1A1

11cos246

(A1)

45.5 ( 0.794 radians) A1

[4 marks]

(c) (i)

1 2 6 21 1 4 62 6 1 2

(M1)

410 4 10 3

3

i j k A1

continued…


Question 13 continued (ii) METHOD 1

let P be the intersection of 1L and 3L

let Q be the intersection of 2L and 3L

1OP 2

2 2

1 2OQ 2

4 6

M1

therefore

2PQ OQ OP

2 6 2

M1A1

2 410

2 6 2 3t

M1

2 4 010 0

6 2 3 2

ttt

solving simultaneously (M1)

32 28 (0.256), ( 0.224)

125 125 A1

Note: Award A1 for either correct or .

EITHER

therefore

931251 0.744282OP 2 2.256125

2 2 2.512314125

A1

therefore 3 3

0.744 4: 2.256 10

2.512 3L

r A1

continued…


Question 13 continued OR

therefore

691251 2 0.552222OQ 2 1.776125

4 6 2.656332125

A1

therefore 3 3

0.552 4: 1.776 10

2.656 3L

r A1

Note: Allow position vector(s) to be expressed in decimal or

fractional form.

[10 marks] METHOD 2

3 3

410

3

aL : r b t

c

forming two equations as intersections with 1L and 2L

1

4 1 110 2 1

3 2 2

ab tc

2

4 1 210 2 1

3 4 6

ab tc

M1A1A1

Note: Only award M1A1A1 if two different parameters 1 2t ,t used.

attempting to solve simultaneously M1

32 28 (0.256), ( 0.224)

125 125 A1

Note: Award A1 for either correct or .

continued…


Question 13 continued EITHER

0 5521 7762 656

a .b .c .

A1

therefore 3 3

0.552 4: 1.776 10

2.656 3L t

r A1A1

OR

0 7442 2562 512

a .b .c .

A1

therefore 3 3

0.744 4: 2.256 10

2.512 3L t

r A1A1

Note: Allow position vector(s) to be expressed in decimal or fractional form.

Total [18 marks]